Đáp án:
25. B
26. \({u_L} = 400\cos \left( {100\pi t + \frac{\pi }{4}} \right)\left( V \right)\)
Giải thích các bước giải:
25.
Trở kháng
\[\begin{array}{l}
{Z_L} = L\omega = \frac{2}{\pi }.100\pi = 200\\
{Z_C} = \frac{1}{{C\omega }} = \frac{1}{{\frac{{{{10}^{ - 4}}}}{\pi }.100\pi }} = 100\\
Z = \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} = \sqrt {{{100}^2} + {{\left( {200 - 100} \right)}^2}} = 100\sqrt 2
\end{array}\]
Ta có:
\[\begin{array}{l}
{I_0} = \frac{{{U_0}}}{Z} = \frac{{200\sqrt 2 }}{{100\sqrt 2 }} = 2A\\
\tan \varphi = \frac{{{Z_L} - {Z_C}}}{R} = \frac{{200 - 100}}{{100}} = 1 \Rightarrow \varphi = \frac{\pi }{4}\\
\varphi = {\varphi _u} - {\varphi _i} \Rightarrow \frac{\pi }{4} = 0 - {\varphi _i} \Rightarrow {\varphi _i} = - \frac{\pi }{4}\\
i = 2\cos \left( {100\pi t - \frac{\pi }{4}} \right)\left( A \right)
\end{array}\]
26.
Trở kháng
\[\begin{array}{l}
{Z_L} = L\omega = \frac{2}{\pi }.100\pi = 200\\
{Z_C} = \frac{1}{{C\omega }} = \frac{1}{{\frac{{{{10}^{ - 4}}}}{\pi }.100\pi }} = 100\\
Z = \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} = \sqrt {{{100}^2} + {{\left( {200 - 100} \right)}^2}} = 100\sqrt 2
\end{array}\]
Ta có:
\[\begin{array}{l}
{I_0} = \frac{{{U_0}}}{Z} = \frac{{200\sqrt 2 }}{{100\sqrt 2 }} = 2A\\
{U_{0L}} = {I_0}.{Z_L} = 2.200 = 400V\\
\tan \varphi = \frac{{{Z_L} - {Z_C}}}{R} = \frac{{200 - 100}}{{100}} = 1 \Rightarrow \varphi = \frac{\pi }{4}\\
\varphi = {\varphi _u} - {\varphi _i} \Rightarrow \frac{\pi }{4} = 0 - {\varphi _i} \Rightarrow {\varphi _i} = - \frac{\pi }{4}\\
{\varphi _{{u_L}}} = {\varphi _i} + \frac{\pi }{2} = \frac{\pi }{4}\\
{u_L} = 400\cos \left( {100\pi t + \frac{\pi }{4}} \right)\left( V \right)
\end{array}\]
